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How can these equations be true?

one and two equals twelve

two and three equals forty six

Then what's three and four? (It's not 102)

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    $\begingroup$ When someone posts an answer the works fine (102), you can't just edit to say it's not the answer. It means the puzzle didn't work. Two data points is way too few; there's lots of reasonable functions that can be fit to them. $\endgroup$ – xnor Apr 15 '15 at 21:20
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I think it's:

$912$

$$\begin{align} 1\,\&\,2 &= (1\times1)(2\times1) \implies 12\\[5pt] 2\,\&\,3 &= (2\times2)(2\times3) \implies 46 \end{align}$$

$$3\,\&\,4 = (3\times3)(3\times4) \implies 912$$ You multiply the first number to itself as the first set of digits, then multiply the first number to the second for the second set of digits. Then put them together into one number.

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  • $\begingroup$ i agree, this was exactly what i was thinking +1 $\endgroup$ – Vincent Apr 15 '15 at 20:37
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Three and four is

102.

The algorithm is:

put the two given digits together to form a 2-digit number, and then multiply by the first of them.

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  • $\begingroup$ Partially right but it's not the correct answer .. $\endgroup$ – user37421 Apr 15 '15 at 20:28
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I note the numbers are spelled out. So, while rand al'thor's approach is probably correct, the intended answer might be 'one hundred two'

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  • $\begingroup$ If that's true, then the title is wrong since it uses numbers for 12 and 46. $\endgroup$ – pacoverflow Apr 15 '15 at 20:33
  • $\begingroup$ Well, I'll be. It does. If only I had scrolled up one more notch. $\endgroup$ – Ayefork Apr 15 '15 at 20:34
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102 is also an answer.

This is how:

$12=1^2\cdot10+1\cdot2$

$46=2^2\cdot10+2\cdot3$

$102=3^2\cdot10+3\cdot4$

according to this operator:

$O(x,y)=x^2\cdot10+x\cdot y$

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  • $\begingroup$ Yes but the question was not complete .. so I added "it's not 102" $\endgroup$ – user37421 Apr 15 '15 at 20:52
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I agree that the accepted answer satisfies the criteria, though additionally it could be:

2712 -> $3^3=27$ & $3\cdot4=12$

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